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3 years ago

Round 824.0472 to the nearest hundredth.

Mathematics

1 answer:

Gelneren [198K]3 years ago

4 0

Answer:

824.05

Step-by-step explanation:

the hundreths place is two spots behind the decimal, there is a four. Look at the number next to it is a 7 so you round up

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The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For

Sergeeva-Olga [200]

Answer:

Required solution gives series (a) divergent, (b) convergent, (c) divergent.

Step-by-step explanation:

(a) Given,

$\sum_{n\to 0}^{\infty}\frac{2^n}{9^{2n}+1}$

To applying limit comparison test, let $a_n=\frac{2^n}{9^{2n}+1}$ and $b_n=\frac{9^{2n}}{2^n}$ . Then,

$\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}(1+\frac{1}{9^{2n}})=1>0$

Because of the existance of limit and the series $\frac{9^{2n}}{2^n}$ is divergent since $\frac{9^{2n}}{2^n}=(\frac{9^2}{2})^n$ where $\frac{81}{2}>1$ , given series is divergent.

(b) Given,

$\sum_{n\to 1}^{\infty}(\frac{7^n}{7^n+4})$

Again to apply limit comparison test let $a_n=\frac{7^n}{7^n+4}$ and $b_n=\frac{1}{7^n}$ we get,

$\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{1}{7^n+4}=0$

Since $\lim_{n\to \infty} \frac{1}{7^n}=0$ is convergent, by comparison test, given series is convergent.

$\sum_{n\to 1}^{\infty}\frac{5^n+2^n}{6^n}= \sum_{n\to 1}^{\infty}(\frac{5}{6})^n+\sum_{n\to 1}^{\infty}(\frac{1}{3})^n$ . Now applying Cauchy Root test on last two series, we will get,

\lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
\lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2

Therefore,

$\lim_{n\to \infty}\frac{5^n+2^n}{6^n}=L_1+L_2=1.16>1$

Hence by Cauchy root test given series is divergent.

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4 years ago